- Each of these triangles has a point U = midpoint of GH. The Fermat

line goes through this point, so either the U points of the three

triangles involved are colinear, or they are the same.

Steve

On Dec 7, 2005, at 9:22 AM, Wilson Stothers wrote:

> Dear All,

>

> We have the following fact :

>

> If F is a Fermat Point of triangle ABC, then F is also a

> Fermat Point of its Cevian, and of its Anticevian Triangle.

>

> A couple of questions :

>

> Is this well-known? I'd appreciate a reference.

>

> Apart from G, are there other centres with this property?

>

> I also observe from Cabri that, if F is a Fermat point,

> and if ABC and the Cevian and Anticevian Triangles have

> Fermat Points {F,F2},{F,F3},{F,F4}, respectively, then

> (a) the points F,F2,F3,F4 are concyclic,

> (b) the circle also contains the isogonal conjugates of

> F2, F3, F4 with respect to their triangles,

> (c) the point F and its isogonal conjugates in the three

> triangles are collinear,

> (d) the Euler Lines of the three triangles are parallel,

> this is an easy consequence of (c).

>

> Of course the isogonal conjugates in (b) and (c) are the

> Isodynamic Points of their triangles. Thus, if F = X13,

> then the circle is that through X13, X14 and X16, and

> the line is X13X15. Note that F need not be X13 for the

> other two triangles, but is a Fermat Point.

>

> I suspect that (c) characterizes the Fermat Points, and

> perhaps that (d) also does.

>

> I'd be glad of any proofs or comments on these. I think

> that a circle with seven points is unusual. The centre

> might be interesting, it is on X115X125 as this is the

> perpendicular bisector of X13X14.

>

> Seasonal Greetings

>

> Wilson

>

>

>

>

>

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> - WILSON SOTHERS CIRCLE<q> I also observe from Cabri that, if F is a Fermat point, and if ABC and the Cevian and Anticevian Triangles have

On Dec 7, 2005 Wilson Sothers posted to Hyacinthos (1):Fermat Points {F,F2},{F,F3},{F,F4}, respectively, then

(a) the points F,F2,F3,F4 are concyclic,

(b) the circle also contains the isogonal conjugates of F2, F3, F4 with respect to their triangles,

(c) the point F and its isogonal conjugates in the three triangles are collinear,

(d) the Euler Lines of the three triangles are parallel, this is an easy consequence of (c). </q>On Aug 30, 2016, Tran Quang Hung posted (2):<q>Let A'B'C' be cevian triangle of F1 and A''B''C'' be anticevian triangle of F1 wrt ABC.F' and F'' are second Fermat points of triangles A'B'C' and A''B''C''.Then four point F1,F2,F',F'' are concyclic on circle (w)So if we construct more anticevian of F1 wrt A''B''C''.... then the second Fermat points of new triangles lie on (w), too.Is this a new circle ? </q>Note:As the author of the page, says in the Foreword:

TTW started taking shape in the 2002 Spring originated from a wide collection of previous notes, a great number of them borrowed from the Hyacinthos mail list and, after, of its follower Anopolis. Let me thank all its members.

http://www.xtec.cat/~qcastell/ttw/ttweng/forew1.htmlC. Lozada posted the center of the circle (4):

<q>The center of the circle is X(9201)= X(14)-OF-2nd-PARRY-TRIANGLE. It passes through ETC's 13,14,15,16,5623 <q>(1)

https://groups.yahoo.com/neo/groups/hyacinthos/conversations/messages/11798

(2)

https://groups.yahoo.com/neo/groups/AdvancedPlaneGeometry/conversations/messages/3409

(3)

https://groups.yahoo.com/neo/groups/AdvancedPlaneGeometry/conversations/messages/3410

(4)

https://groups.yahoo.com/neo/groups/AdvancedPlaneGeometry/conversations/messages/3410APH